Question

I'm confused a little...
How does the integral of secxtanx = secx ?

Answer 1

it doesnt ._.

Answer 2

Really? XD lol
Then what is is?

Answer 3

it*

Answer 4

sec(theta)=\[\frac{ 1 }{ \cos*theta}\]

Answer 5

Right

Answer 6

if y = sec(x), then dy/dx = sec(x)*tan(x)

take this in reverse to get \[\Large \int(\sec(x)\tan(x))dx = \sec(x) + C\]

Answer 7

Ohhh right! That makes sense!!

Answer 8

if you had no idea that dy/dx = sec(x)*tan(x)

then you can convert sec(x)*tan(x) into (1/cos)*(sin/cos) which becomes sin/(cos^2)

from there you use u-sub

u = cos(x)
du = -sin(x)dx

Answer 9

that's one of the standard derivative definitions...
The derivative of secx is secxtanx
the antiderivative of secxtanx is secx.

Answer 10

Yeah I would've had to go through it with u-sub... i didn't know that dy/dx of secx = secxtanx

Answer 11

Right! Thanks you guys! That really helped!

Answer 12

you're welcome

Answer 13

:)